Provided that these integrals can be evaluated and that they're not too difficult to do so, then we can obtain solutions for the separable differential equation.

We will now look at some examples of solving separable differential equations. In these examples, we will concern ourselves with determining the Interval of Validity, which is the largest interval for which our solution is valid that contains the initial condition given.

We note that this solution is valid provided that $y \neq 0$ (so that the denominator of the differential equation is nonzero) and $\frac{x^2 + 1}{2} ≥ 0$ (so that we don't have the square root of a negative number). We see that $\frac{x^2 + 1}{2} ≥ \frac{1}{2} > 0$ for all $x \in \mathbb{R}$ and so $y = - \sqrt{\frac{x^2 + 1}{2}} \neq 0$.

We note that $-2 ≤ x ≤ 2$ so that the square root above is not negative, and so $2 ≥ \sqrt{4 - x^2} ≥ 0$. So to achieve the initial condition $y(0) = 1$ we must have that $y = 3 - \sqrt{4 - x^2}$. So our interval of validity is $[-2, 2]$.

To prevent the denominator from equalling zero in our solution, we have that $x \neq -2$ and $x \neq 3$. Note that $x = 0$ provides out initial value $y(0) = -\frac{1}{6}$ and so our interval of validity must contain $x = 0$ so $(-2, 3)$ is our interval of validity.